When the area grows
Steam pipes in a factory or a dairy are wrapped in thick lagging to keep the heat in. Heat flows from the hot pipe outward through the lagging, along the radius. The slab formula \(H = kA\,\Delta T/L\) cannot be used here, because there is no single area \(A\). Near the pipe the heat crosses a small cylinder; further out it crosses a larger one.
The way out is to cut the lagging into many thin cylindrical shells. Take one shell of radius \(r\), thickness \(dr\) and length \(L\). Its area is \(2\pi rL\), and it is thin enough to treat as a slab. In the steady state the same heat current \(H\) crosses every shell, since no heat collects anywhere inside the lagging.
\[ H = -k\,(2\pi rL)\,\dfrac{dT}{dr} \]
Why. Fourier's law across the thin shell of radius \(r\), whose area is \(2\pi rL\).
\[ dT = -\dfrac{H}{2\pi kL}\,\dfrac{dr}{r} \]
Why. \(H\) is the same for every shell, so separate the variables.
\[ T_2 - T_1 = -\dfrac{H}{2\pi kL}\,\ln\dfrac{r_2}{r_1} \]
Why. integrate from the inner surface, radius \(r_1\) at \(T_1\), to the outer surface, radius \(r_2\) at \(T_2\).
\[ H = \dfrac{2\pi kL\,(T_1 - T_2)}{\ln(r_2/r_1)} \]
Read it. solve for \(H\).
A steam pipe of outer radius \(5\,\text{cm}\) has its surface at \(120\,°\text{C}\). It is wrapped in lagging of \(k = 0.04\,\text{W/(m K)}\) out to a radius of \(10\,\text{cm}\), whose outer surface is at \(20\,°\text{C}\). Find the heat lost per metre of pipe, and the thermal resistance of the lagging per metre. Take \(\ln 2 = 0.693\).
\(H = 2\pi kL\,\Delta T/\ln(r_2/r_1) = 2\pi \times 0.04 \times 1 \times 100/\ln 2 = 25.13/0.693 = 36.3\,\text{W}\).
\(R_\text{th} = \ln 2/(2\pi \times 0.04 \times 1) = 0.693/0.251 = 2.76\,\text{K/W}\), and \(100/2.76 = 36.3\,\text{W}\), as before.
Only the ratio of the radii enters, not the thickness of 5 cm by itself.
Try the slider in the figure above. With lagging to 10 cm the loss is 36.3 W per metre. Doubling the outer radius to 20 cm triples the lagging, from 5 cm to 15 cm thick. Yet the loss only halves, to 18.1 W. The ratio \(r_2/r_1\) doubles from 2 to 4, and \(\ln 4 = 2\ln 2\). Going on to 25 cm saves only 2.5 W more. Each extra centimetre of lagging buys less.
The lagging of the example is made thicker, out to a radius of \(15\,\text{cm}\), with the surfaces still at \(120\,°\text{C}\) and \(20\,°\text{C}\). Find the heat lost per metre. Take \(\ln 3 = 1.099\).
The outer radius of the lagging on a pipe is increased from \(2r_1\) to \(4r_1\), with the surface temperatures unchanged. The heat current becomes
A spherical shell
A round vessel wrapped in lagging, or a hot ball inside a thick shell, sends heat outward through spherical shells. A thin shell of radius \(r\) has area \(4\pi r^2\), which grows faster with \(r\) than a cylinder's. The same method works. Fill in the missing step below.
\[ H = -k\,(4\pi r^2)\,\dfrac{dT}{dr} \]
Why. Fourier's law across a thin spherical shell of radius \(r\).
\[ dT = -\dfrac{H}{4\pi k}\,\dfrac{dr}{r^2} \]
Why. \(H\) is the same at every radius, so separate the variables.
?
Your turn. integrate from \(r_1\) to \(r_2\), using \(\int dr/r^2 = -1/r\). Which line is it?
\[ H = \dfrac{4\pi k r_1 r_2\,(T_1 - T_2)}{r_2 - r_1} \]
Read it. \(1/r_1 - 1/r_2 = (r_2 - r_1)/(r_1r_2)\); solve for \(H\).
A round vessel of hot milk in a dairy has an outer radius of \(10\,\text{cm}\), with its surface at \(90\,°\text{C}\). It is covered by a spherical layer of lagging, \(k = 0.04\,\text{W/(m K)}\), out to a radius of \(20\,\text{cm}\), whose outer surface is at \(30\,°\text{C}\). Find the heat lost through the lagging.
\(H = 4\pi k r_1 r_2\,\Delta T/(r_2 - r_1) = 4\pi \times 0.04 \times 0.1 \times 0.2 \times 60/0.1\).
\(4\pi \times 0.04 = 0.503\); times \(0.02\) gives \(0.01005\); times \(60/0.1 = 600\) gives \(H = 6.03\,\text{W}\).
Two limits of the sphere formula are worth knowing. If the shell is thin, \(r_2 - r_1 = d\) is small and \(r_1r_2 \approx r_1^2\). Then \(H \approx 4\pi r_1^2\,k\,\Delta T/d\), the slab formula with area \(4\pi r_1^2\). If instead the shell is made very thick, \(r_2 \to \infty\), then \(r_1r_2/(r_2 - r_1) \to r_1\). So \(H\) does not fall to zero; it falls only to \(4\pi kr_1\,\Delta T\).
For the vessel, that limit is \(4\pi \times 0.04 \times 0.1 \times 60 = 3.02\) W. Lagging to 20 cm already gives 6.03 W, twice the limit, and no amount of extra lagging can halve it again. The outer layers have huge areas, so they add almost no resistance. Nearly all the resistance lies in the first few centimetres around the vessel.
A spherical shell of lagging has inner radius \(10\,\text{cm}\) and outer radius \(10.5\,\text{cm}\), with \(k = 0.04\,\text{W/(m K)}\) and \(60\,\text{K}\) across it. Find the heat current from the exact sphere formula.
The temperature inside the lagging
In a slab the steady temperature falls in a straight line. Inside lagging it does not. The same heat current must cross every shell. Near the inner surface the area is smallest, so the heat needs a steep gradient to squeeze through. Further out the area is larger and a gentle gradient is enough. For the pipe, \(dT/dr = -H/(2\pi kLr)\) falls as \(1/r\), and the temperature falls as \(\ln r\).
Fig 5.1 compares the three shapes. Halfway through the wall, the slab is at 50 °C. The pipe's lagging is at \(100\,(1 - \ln 1.5/\ln 2) = 41.5\) °C, and the sphere's at 33.3 °C. In the steam pipe of the example, the lagging at 7.5 cm is at 61.5 °C, not the 70 °C of a straight line. At the inner surface the gradient is \(r_2/r_1\) times that at the outer surface for a pipe, and \((r_2/r_1)^2\) times for a sphere.
Integrating only as far as a radius \(r\), instead of all the way to \(r_2\), gives the temperature at any point of the shell. For a pipe the fraction of the drop already made at radius \(r\) is \(\ln(r/r_1)/\ln(r_2/r_1)\). For a sphere it is \((1/r_1 - 1/r)/(1/r_1 - 1/r_2)\). In the lagged milk vessel of the example, at \(r = 15\,\text{cm}\) the sphere fraction is \((1/10 - 1/15)\) divided by \((1/10 - 1/20)\), which is \(2/3\). So two thirds of the 60 K drop is already made, and the lagging there is at \(90 - 40 = 50\,°\text{C}\).
In the steady state, where is the temperature gradient largest inside the lagging of a steam pipe?
A spherical shell has inner radius \(5\,\text{cm}\) and outer radius \(15\,\text{cm}\). In the steady state, how many times steeper is the temperature gradient at the inner surface than at the outer surface?
In a flat slab the steady temperature falls in a straight line. Explain why, inside the lagging of a pipe, it is not a straight line against the radius.
In the steady state the same heat current crosses every cylindrical surface in the lagging. The area of such a surface, \(2\pi rL\), grows with the radius. Through a small area the heat needs a steep gradient, and through a large area only a gentle one. So the gradient falls as \(1/r\) and the temperature falls as \(\ln r\), steeply near the pipe and gently further out.
A sugar mill runs long pipes of steam from its boilers to its heaters and evaporators. With the lagging of the example, each metre still loses about 36 W, or \(36 \times 24/1000 \approx 0.87\,\text{kWh}\) a day. Along 500 m of pipe that is about 435 kWh a day. Without lagging the loss would be many times larger. Engineers choose the thickness where the fuel saved by one more centimetre no longer pays for it.
- Using \(H = kA\,\Delta T/L\) for lagging with the area of the pipe.The area grows with the radius; use \(2\pi kL\,\Delta T/\ln(r_2/r_1)\) for a cylinder and \(4\pi kr_1r_2\,\Delta T/(r_2 - r_1)\) for a sphere.
- Thinking that doubling the thickness of lagging halves the loss.For a cylinder the loss goes as \(1/\ln(r_2/r_1)\); it halves only when \(r_2/r_1\) is squared.
- Drawing a straight temperature profile through lagging.It falls as \(\ln r\) in a cylinder and as \(1/r\) in a sphere, steepest at the inner surface.